A system releases 125ky of heat, & 104 Ky Of Work
is done on it. Calculate LE

Problemas de Termodinámica. Primera Ley.

1) a) ¿Cuál es el cambio de energía interna, cuando

un sistema pasa del estado a al b a lo largo de

la transformación acb recibe una cantidad de

calor de 20000 cal y realiza 7.500 cal de

trabajo?

b) ¿Cuánto calor recibe el sistema a lo largo de

la transformación adb, si el trabajo realizado

es de 2500 cal?

c) Cuando el sistema vuelve de b hacia a, a lo

largo de la transformación en forma de curva,

el trabajo hecho sobre el sistema es de 5000

cal. ¿Cuánto calor absorbe o libera el sistema?

d) Si Ua = 0 y Ud = 10000 cal hállese el calor

absorbido en los procesos ad y db.

Nota: no hace falta ningún dato de P y V para resolver

el problema

Respuesta

a) El camino acb:

qacb = 20000 cal; wcb = –7500 cal; wac = 0

Uab = qacb + wacb = 20000 – 7500 = 12500 cal.

b) El camino adb:

Uab = qadb + wadb  wadb = –2500 cal;

12500 = qadb – 2500  qadb = 15000 cal

c) El camino ba:

Uba = –12500 cal  wba = 5000 cal;

Uba = qba + wba  –12500 = qba + 5000

qba = –17500 cal

d) El camino ad:

Uad = qad + wad  Uad = Ud – Ua = 10000 – 0

Uad = 10000 cal

wad = wadb = – 2500 cal; ya que wbd = 0

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.